DM ac: Redox titrations & equations Flashcards
Explain the basic premise of manganate(VII) titrations.
- Acidified MnO4-(aq) oxidises Fe2+ ions to Fe3+ ions
- Known volume of Fe2+ is titrated with KMnO4 of known concentration
- MnO4-(aq) is purple. No indicator is required since end point, when MnO4-(aq) is in excess, is when pink colour persists
Potassium manganate(VII) oxidises iron(II) ions to iron(III) ions in acidic solution. Write the 2 half-equations and the full equation for this reaction.
Fe2+ → Fe3+ + e-
MnO4- + 5e- + 8H+ → Mn2+ + 4H2O
KMnO4- + 5Fe2+ + 8H+ → K+ + 5Fe3+ + Mn2+ + 4H2O
The concentration of a solution of Fe2+ ions can be found by titration with a standard potassium manganate(VII) solution.
- Describe how the titration would be carried out.
- State how the end-point can be recognised and explain why an indicator does not need to be added.
- Rinse + fill burette with standard MnO4-(aq)
- Use graduated pipette to transfer Fe2+(aq) to conical flask
- Add excess H2SO4 to flask
- Titrate until pink colour remains due to excess MnO4-(aq); no indicator required
- Repeat to obtain 3 concordant results within 0.1 cm3
Don’t use HCl since it is easily oxidised to toxic Cl2
The end point of a manganate(VII) titration is taken as the when the pink colour of the solution in the conical flask persists for 10 seconds. Suggest why it might fade.
- End point was not actually reached since reaction is slow
- Reaction of MnO4- with water
- Reaction of MnO4- with Mn+ (forming MnO2)
25 cm3 of an iron(II) sulfate solution, in excess acid, requires an average titre of 17.3 cm3 0.0200 mol dm-3 potassium manganate(VII) solution to reach a permanent pink colour.
- What is the concentration of Fe2+(aq) in the solution?
- The original iron(II) sulfate solution was obtained by dissolving a sample of 27.50 g in excess sulfuric acid and making it up to 2000 cm3. What was the percentage of iron in the ore? Mr of Fe = 55.8, S = 32.1, O = 16
5Fe2+(aq) + MnO4-(aq) + 8H+(aq) → 5Fe3+(aq) + Mn2+(aq) + 4H2O(l)
Mol MnO4- = 0.02 x 17.3 x 10-3 = 3.46 x 10-4
Mol Fe2+ = 5 x 3.46 x 10-4 = 1.73 x 10-3
Conc Fe2+ in original solution = 1.73 x 10-3 / (25 x 10-3) = 0.069 mol dm-3
Mol Fe2+ in 2000 cm3 = 0.069 x 2 = 0.138 mol
Mass FeSO4 in 2000 cm3 = 0.138 x 151.9 = 20.96 g
% mass = 20.96 x 100 / 27.5 = 76%
As an alternative to potassium manganate(VII) solution, potassium dichromate solution may be used in a redox titration to determine the concentration of a solution of Fe2+.
Explain the basic premise of this.
- Acidified potassium dichromate(VI) oxidises Fe2+ to Fe3+
- Known volume of Fe2+(aq) titrated with Cr2O72-(aq) of known concentration
- Cr2O72-(aq) is orange, so end point, when Cr2O72-(aq) is in excess, is when orange colour persists. No indicator required
Potassium dichromate(VI) oxidises iron(II) ions to iron(III) ions in acidic solution. Write the 2 half-equations and the full equation for this reaction.
Cr: ox state +6 → +3, so gain of 3 electrons. But 2 Cr atoms are involved, so equation involves 6 electrons
Cr2O72- + 14H+ + 6e- → 2Cr3+ + 7H2O
Fe2+ → Fe3+ + e-
6Fe2+ + Cr2O72- + 14H+ → 6Fe3+ + 2Cr3+ + 7H2O
![](https://s3.amazonaws.com/brainscape-prod/system/cm/343/090/191/q_image_thumb.png?1618685172)
Mol KMnO4 = mol MnO4- = 0.05 x 18.4 x 10-3 = 23/25,000
Mol C2O42- in 25 cm3 = 23/25,000 x 5/2 = 23/10,000
Mol C2O42- in 100 cm3 = 23/2500
So mol K2C2O4 • H2O = 23/2500
Mr K2C2O4 • H2O = 184.2
Mass K2C2O4 • H2O = 23/2500 x 184.2 = 1.69 g
![](https://s3.amazonaws.com/brainscape-prod/system/cm/343/090/192/q_image_thumb.png?1618685172)
Mol C2O42- = 0.05 x 10 x 10-3 = 1/2000
Mol MnO4- = 1/2000 x 2/5 = 1/5000
Conc MnO42- = 1/5000 / (26 x 10-3) = 0.00769 mol dm-3
![](https://s3.amazonaws.com/brainscape-prod/system/cm/343/090/193/q_image_thumb.png?1618685172)
Mol S2O32- = 0.2 x 20.5 x 10-3 = 41/10,000
Mol Cu2+ in 25 cm3 = 41/10,000
Mol Cu2+ in 250 cm3 = 41/10,000 x 250/25 = 41/1000
Mass Cu2+ in coin = 41/1000 x 63.5 = 2.6035 g
% mass Cu = 2.6035 x 100 / 3.47 = 75.0 %
When zinc is added to copper(II) sulfate solution, it displaces the copper.
Suggest two changes which would be observed.
- Blue solution becomes paler
- Copper metal deposits on zinc
Balance the following equation:
Sn + HNO3 → SnO2 + NO2 + H2O
Sn + HNO3 → SnO2 + NO2 + H2O
Balance ox states:
Sn: 0 → 4 (gains 4 e-)
N: 5 → 4 (loses 1 e-)
Sn + 4HNO3 → SnO2 + 4NO2 + H2O
No charge to balance
Balance rest
O: 12 → 11
H: 4 → 2
Sn + 4HNO3 → SnO2 + 4NO2 + 2H2O
Balance the following equation:
MnO4- + Fe2+ + H+→ Mn2+ + Fe3+ + H2O
MnO4- + Fe2+ + H+→ Mn2+ + Fe3+ + H2O
Balance ox states:
Mn: +7 → +2 (gains 5 e-)
Fe: +2→ +3 (loses 1 e-)
MnO4- + 5Fe2+ + H+→ Mn2+ + 5Fe3+ + H2O
Balance charge
As is: +10 → +17
Only alter species not involved in redox:
MnO4- + 5Fe2+ + 8H+→ Mn2+ + 5Fe3+ + 4H2O
No rest to balance
Write a half equation to represent the change when an aqueous solution of chlorate(I), ClO-, ions is reduced, forming chloride ions and hydroxide ions.
ClO- + 2e- + H2O → Cl- + 2OH-
Hydrogen peroxide solution was added to a test tube containing sodium chlorate(I) solution.
- Decide which species are reduced and oxidised and deduce the two half-equations
- Write the redox requation
- Use this to suggest what is observed
Reduction:
ClO- + 2e- + H2O → Cl- + 2OH-
+1 to -1: gain of 2
Oxidation:
H2O2 → O2 + 2H+ + 2e-
-1 to 0: loss of 1
Overall:
ClO- + H2O + H2O2 → Cl- + 2OH- + O2 + 2H+
Observation: Fizzing as oxygen gas forms